•ffl-[5(I)+9]1/2 - Lower Moreland Township School ... 4.3 The Chain Rule Thus, q2 = 7500. P'(q) =...

Section 4.3 The Chain Rule

Thus, q2 = 7500.

P'(q) = 8000 - 2q2= 8000 - 2(7500)= 8000 - 15,000

= -7000

When the price is $5000, the marginal profit is-$7000.

59. P{x) = 2x2 + 1; x = f(a) = 3a + 2

P[/(a)]=2(3a +2)2 + l= 2(9a2 + 12a + 4) + 1= 18a2 + 24a + 9

60. (a) A(r) = irr2r(t) = t2

A[r(t)]=*[t2)2= *{tA)= nt4

This function represents the area of the oil slickas a function of time t after the beginning of theleak.

(b) DtA[r(t)} = 4?rt3DtA[r{l00)\ = 4tt(100)3

= 4,000,000tt

At 100 minutes the area of the spill is changing atthe rate of 4,000,000tt ft2/min.

61. (a) r(t) = 2t; A(r) = 7rr2

A[r(t)}=n(2t)2= 4irt2

A = Airt2 gives the area of the pollution in termsof the time since the pollutants were first emitted.

(b) DtA[r(t)} = 8irtDtA[r(A)] = 8tt(4) = 32tt

At 12 P.M., the area of pollution is changing atthe rate of 32tt mi2/hr.

62. N(t) = 2t{5t + 9)l/2+12

N'(t) = (2t) i(5t +9)"1/2(5)

+ 2(5* + 9)1'2 + 0= 5*(5* + 9)-1/2 + 2(5i+ 9)1/2= (5i + 9)"1/2[5< + 2(5/, + 9)]= (5< + 9)"l/2(15«+ 18)

15* + 18

" (5i + 9)!/2

(a) jV(U]- [5{0)+9]1/218

91/2= 6

(b) N•ffl-15 (I)+ 18

[5(I)+9]1/221+8

(7 + 9)V2

39

(16)1/2

!-•15(8) +18

[5(8) + 9]!/2

120 + 18

(49)!/2

= H? « 19.717

63. C(t) = i(2i +1)-1/2

C"(t) =i(-i)(2i+l)-3/2(2)=-i(2f+l)-3/2

(a) C"(0) =-i[2(0) +l]"3/2__1

2

= -0.5

(b) C"(4) =-i[2(4) +l]"3/2

=-^(9)"3/2-1 1

=f=9.754

(c) JV'(8) =

2 (v/9)3_ _J_~~54

« -0.02

(c) C"(7.5) =-i[2(7.5) +l]"3/2

2V(v^6)3J1

128

-0.008

263

264

(d) C is always decreasing because

C" = -£(2f+l)-3/2

is always negative for t > 0.(The amount of calcium in the bloodstream willcontinue to decrease over time.)

64. (a) R(Q)=q(c-^J

R'(Q) = Q

0 , (c Q\'2e(c-f)'/2 +̂ 3^

+

(b) #(Q) = - Q

If Q = 87 and C = 59, then

R'(Q)87\

59 -t)1/2

87

**(c-f)

87

6(59-f)1/2

= (30)1/2 _

= 5.48

6(30)1/2

87

1/2

1/2

32.88

= 5.48 - 2.65

= 2.83.

(c) Because R'{Q) is positive, the patient's sensitivity to the drug is increasing.

65. V(r) =^Trr3, S(r) =4irr2,r(t) =6- ^t

(a) r(t) = 0 when 6 - — t = 0;

t =iM=34mln.

Chapter 4 CALCULATING THE DERIVATIVE

... dV A 2 dS dr 3(b)-=4.r2,-=8.r,- = --

dV dV dr 12

dt dr ' di 17 ™

•H-b)dS dS dr 24

dt ~ dr ' dt ~ 177rr

24 (r 3/\"it'^-itVWhcii t = 17,

dV

dt

12

= -17"6-|(17)

108 ,, .= ——-TV mitr/imn

17 '

dS

dt

24

= -17* 8-£(17)^72

= —-7r mm2/min17

At t = 17 minutes, the volume is decreasing by•^~7r mm3 per minute and the surface area is decreasing by y=;7r mm2 per minute.

4.4 Derivatives of ExponentialFunctions

1. y = e4cLet g(x) = 4x,with g'(x) = 4.

dt/

d.r= 4e4;i:

2. t/ = e_2xLet g(x) = -2.T,

with g'(x) = -2.

^ = -2e"2*dx

3. y = -8e3;r

dydx

= -8(3e3x) = -24e3a

4. t/ = 1.2e5;r

dy

dx= 1.2(5e5x) = 6e5a

Section 4.4 Derivatives of Exponential Functions

5. y = -16e2x+1g(x) = 2x + 1g'{x) = 2

dydx

= -16(2e2x+1) = -32e2x+1

-0.3a:y = -4e

g =-4(-0.3e—)= 1.2e-°-3x

7. y = exg(x) = x2

g'(x) = 2x

* = 2xe**dx

8. y = e-*2g{x) = -x2

g'(x) = -2x

* = -2ze-*'dx

9.

10.

y = 3e2xg(x) = 2x2

g'{x) = 4x

dy

dx= 3(4xe2x >

= 12.Te2a;2

y = -5e4xg(x) = 4x3

g'(x) = Ylx2

dydx

= (-5)(12x2)e

= -60x2e4x3

4xJ

11. y = 4e2x2-4g{x) = 2x2 - 4

g'{x) = 4x

rfj/ =4[(4ar)crfx

2a.-'-41

= 16a:e2a:2-4

12. y = -3e3x2+5#(x) = 3x2 + 5

tf'(x) = 6xy' = (-3)(6x)e3x2+5

= -18xe3x2+5

13. y = xex

Use the product rule.

dy-j- = xex + ex • 1dx

= ex(x + l)

14. y = x2e 2x

17.

18.

19.

Use the product rule

dy

dx̂=x2(-2e~2x) +2xe-2x

= -2x2e~2x + 2xe~2x= 2x(l-x)e~2x

15. y = (x + 3)2e4x


•£ =(x +3)2(4)e4x +e4x •2{x +3)= 4{x + 3)2e4x + 2(x + 3)e4x= 2(x+ 3)e4x[2(x + 3) + l]= 2(x + 3)(2x + 7)e4x

16. y = (3.t3 - 4x)e"5x


~ = (3.T3 - 4.-r)(-5e-5x) + e-5x(9a;2 - 4)= (-15.-U3 + 20x)e-5x + (9a;2 - 4)e_5x= (-15a;3 + 9x2 + 2Qx - 4)e_5x

x2ex

Use the quotient rule.

dy ex(2x) - x2ex

y =y 2x +1dy _ ex(2x + l)-(2)(ex)dx (2x + l)2

_ e*(2x + 1-2)(2x + 1)2

_ ex(2x - 1)(2a:+ 1)2

dx (e*)2

=

xex{2-x)e2x

=

x{2 - x)ex

ex

ex + e-xy =

dy _ x(ex - e~x) - (ex+ e~x)dx

265

266

20. y =ex — e x

dy _ x(ex - (-l)e-T) - (ex- e~x)(l)dx

xex + xe x — ex + e~x

_ ex(x-l) + e-x(x + 1)

21. p =10,000

9 + 4e~0-2'

dp _ (9+ 4e-°2f) •0-10,000(0 + 4(-0.2)e-°-2t]dt (9 + 4e-°-2<)2

8000e"0-2'

22. p =

(9 + 4e-°-2')2

500

12 + 5e-°-5'

dp = (12+5e-°5<)-0-500[0+5(-0.5)e-°5tldt ~ (12 + 5e-°-5<)2

1250e-°-5/

~ (12 + 5e-°-5t)2

23. f{z) = {2z + e-*2?f'(z) = 2{2z + e~^)1(2 - 2ze~z2)

= 4{2z + e-z2)(l-ze-22)

24. y = 73x+1

Let g(x) = 3x + 1, with g'(x) = 3. Then

dydx

= (ln 7)(73x+1)-3

= 3(ln 7)73x+1

25. y = 4"5x+2

Let g(x) = -5x + 2, with g'{x) = -5. Then

dydx

= (ln4)(4-5x+2)-(-5)

= -5(ln4)4-5x+2

26. y = 3-4x2+2

Let g{x) = x2 + 2, with g'(x) = 2x. Then

dydx

= 3(ln 4)4X +2 •2x

= 6x(ln 4)4xJ+2

27. v = -103x' -4

Let g(x) = 3a;2 - 4, with g'(x) = 6x.

^ =-(lnl0)103x2-4-6xdx

= -6x(103x "4)ln 10


28. s = 2-3^

Let g(t) = y/i, with g'(t) = —T. Then2yt

1

V75-»0»^-O(In 3)3^

~ v^

29. s = 5 • 2v/rr2

Let ff(0 = *- 2, with <?'(0 = 5^. Then

^ =5(ln2)(2v/7^).^L=

_^5Jn_2j2_^!2v/r=l

"•-SriUse the quotient rule and product rule.

dy = (e2( + l)(te' +e'-l)- (tel + 2)(2e2')dt (e2t + l)2

(e2< + l)(te'+e')-(fef+2)(2e2')(e2' + l)2

_ fe3' + e3t + te' + el - 2te3t - 4e2t(e2t + l)2

-te31 + e3t + te* + e' - 4e2'

(e2' + 1)2

_ (l-Qe3(-4e2* + (l+Qc'{e2t + l)2

t2e2t31. y =

3t£ + e

Use the quotient rule and product rule.

dy _(t + e3t)(2te2t + t2 •2e21) - t2e2t(l + 3e3/)dt ~ (t + e3')2

_ (< + e3t)(2te2t + 2/2c2') - <2e2'(l + 3e3')(t + e3')2

_ (2£2e2' + 2*3e2< + 2te™ + 2<2e5') - (*2e2' + 3t2e(t + e3t)2

_ t2e2t + 2t3e2t + 2fe5t - /-V"(* + e3')2

(2/,3 + t2)e2t + (2t - t2)e5t(t + c3t)2

2a5t'


32. f(x) = ex^3x+2Let g(x) = x\/3x + 2.

g'{x) = 1•v^x + 2+ x (—===== Jv ' \2s/3TT2J

= x/3x + 2 +3x

2y/3x + 2

2(3x + 2) 3x2V3X + 2 2v'3x + 2

9x + 4

2v/3x+"2

33. /(x) = ex2/(x3+2>

Let g(x) =

9'{x) =

x3 + 2

(x3 + 2)(2x)-x2(3x2)(x3 + 2)2

2x4 + 4x - 3x4

(x3 + 2)2

4x —x4

(x3 + 2)2

x(4 - x3)(x3 + 2)2

fix) = e*2/(x3+2> x(4 - x3)(x3 + 2)2

x(4 - x3)ex2/(x3+2)(x3 + 2)2

— *, akt34. y = y0e

ft =Jt[y°ekt] =Vokekt =k{V°ekt) =̂35. Graph

y =

,x+0.0001-e"

0.0001

on a graphing calculator. A good choice for theviewing window is [—1,4] by [-1,16] with Xscl =1, Yscl = 2.

267

If we graph y = ex on the same screen, we see thatthe two graphs coincide. They are close enough to

being identical that they are indistinguishable.By the definition of the derivative, if f(x) = ex,

nx) =lim /(*+*)-/(«>. lim ^-^J v ' /i-»o h h-o h

and h = 0.0001 is very close to 0.

Comparing the two graphs provides graphical evidence that

/'(x)=ex.

36. Graph the function y = ex

I I I I I2 4

Sketch the lines tangent to the graph at x = —1,0, 1, 2.

Estimate the slopes of the tangent lines at thesepoints.

At x = —1 the slope is a little steeper than ^ orapproximately 0.3.

At x = 0 the slope is 1.Ata; = 1the slope isa littlesteeper than | or 2.5.At x = 2 the slope is a little steeper than 7^ or7.3.

Note that e~l « 0.36787944, e° = 1,c1 = e « 2.7182812, and e2 « 7.3890561. Thevalues are close enough to the slopes of the tangentlines to convince us that ^- = ex.

268

37. S(t) = 100- 90e-°-3'S'(t) = -90(-0.3)e-°-3t

= 27e-03t

(a) 5/(l) = 27e-°-3(1)= 27e-°-3

w20

(b) S"(5) = 27e-°-3<5)= 27e"15

«6

(c) As time goes on, the rate of change of sales isdecreasing.

(d) S'(t) = 27e-°-3£ ^ 0, but

lim S'(t) = lira 27e-°-3' = 0.

Although the rate of change of sales never equalszero, it gets closer and closer to zero as t increases.

38. C{x) = v/900-800-1.1-^

C{x) = [900- 800(1.1-*)]1/2

= i[900-800(l.l-iC)]-1/2• [-800(ln l.l)(l.l-x)(-l)]

(400 In 1.1)(1.1-T)C'{x) =

x/900 - 800(1.1"T)

vTooThe marginal cost is $3.81.

W V ' ^900-800(1.1-2°)The marginal cost is $.20.

(c) As x becomes larger and larger, C'{x) approaches zero.

39. A(t) = 10t2 2~lA'(t) = 10t2(\n 2)2-'(-l) + 20* 2~lA'(t) = 10t2-l(-t In 2 + 2)

(a) A'{2) = 10(2)(2"2)(-2 In 2 + 2)«3.07

(b) A'(4) = 10(4)(2"4)(-4 In 2 + 2)« -1.93

(c) Public awareness increased at first and thendecreased.


40. /(/,) = 0.028(3.824)'-2001/'(/,) = 0.028(ln3.824)(3.824)<-2001(l)

= 0.028(ln3.824) (3.824)'"2001

(a) /'(2002) = 0.028(ln3.824)(3.824)2002-2001« 0.144

The instantaneous rate of change in 2002 is 0.144million or 144,000 subscribers per year.

(b) /'(2006) =0.028(ln3.824)(3.824)200C-2001«30.7

The instantaneous rate of change in 2006 is 30.7million or 30,700,000 subscribers per year.

41. y = I00e-O03015t

(a) For t = 0,

(b) For t = 2,

(c) For t = 4,

y = I00e-O03045(°>= 100e°

= 100%.

y = l00e-°-O3O45(2>= lOOe"00609

« 94%.

y = lOOe-0-03045^« 89%.

(d) For t = 6,

// = lOOe"0-03045*6)« 83%.

(e) y' = 100(-0.03045)e-003045'= -3.045e-°03045'

For /, = 0,

y' = -3.045e-°03O45(°)= -3.045.

(f) For t = 2,

y' = -3.045e-°03045(2)« -2.865.

(g) The percent of these cars on the road is decreasing, but at a slower rate as they age.


42. S(t) = SOOOe0-1^0 25t)

Let g(t) = 0.1e025t, withg'{t) = 0.1(e°-25t)(0.25) = 0.025e°-25<

S'(t) = 5000e01(e° 25<)(0.025e°-25')= 125e01(e° 25<)e°-25*

•025t)+0.25t

S'(8) = 125e0A(e°™8)+0-25W= 125e0,(e2)+2«1934

The answer is b.

43. (a) G0 = 0.7, m = 10.3

G{t) = 10-3

G'(5)

1+ (JJ^ _ 1)e-0.03036(10.3)t

10.3

~ 1 + 13.71e-°-3127t

(b) G'{t) = -10.3(1 + 13.71e-°-3127')-2•13.71e-°-3127t(-0.3127)

44.1573051e-°-3127t

~ (1 + 13.71e-°-3127t)21990 when t = 5:

G(5) = 10-31 ' 1+ 13.71e-°-3127(5)

2.66

44.1573051e-°-3127(5>

[1 + 13.71e-°-3127(5)]2

« 0.617

The population in 1990 is 2.66 million and thegrowth rate is 0.617 million per year.

(c) 1995 when t = 10:

10.3G(10) =

1 + 13.71e-°-3127(10)

6.43

44.1573051e-°-3127<10)G'(10) =

G'(15) =

[1 + 13.71e-°-3127(10)]2

« 0.755


(d) 2000 when t = 15:

'15) = 1+ 13.71e-0.3127(15)«9.15

44.1573051e-°-3127(15>

[1 + l3.71e-°-3127(15>]2

« 0.320

269


(e) The rate of growth over time increases for awhile and then gradually decreases to 0.

44. (a) Go = 0.7, m = 13.7

13.7G{t) = 1+ f&J. _ 1)e-0.02352(13.7)t

13.7

1 + 18.57e-°-3222t

(b) G'(t) = -13.7(1 + 18.57e-°3222t)-2•18.57e-°3222t(-0.3222)

_ 81.9705798e-°-3222t~ (1 + 18.57e-°-3222')2

1990 when t = 5:

13.7G(5) =

G'(5) =

1 + 18.57e-°-3222(5)

2.91

81.9705798e-°-3222(5>

[1 + 18.57e-°-3222(5)]2

0.738


(c) 1995 when t = 10:

(10) = 1+ i8.57e-°-3222(10)

G'(10) =

7.87

81.9705798e-°-3222(10>

[1 + 18.57e-°-3222(10)]2

1.079


(d) 2000 when t = 15:

G(15) = —y } 1 + 18.57e-°-3222(15)

G'(15) =

11.94

81.9705798e-°3222(15)

[1 + 18.57e-°-3222(15>]2

w 0.495


(e) The rate of growth over time increases for awhile and then gradually decreases to 0.

270

45. p(t) = 9.865(1.025)'p'(f) = 9.865(lnl.025)(1.025)t

(a) For 1998, t = 18.

p'(18) = 9.865(lnl.025)(1.025)18= 0.380

The instantaneous rate of growth is 380,000peopleper year.

(b) For 2006, t = 26.

p'(26) = 9.865(ln 1.025)(1.025)26= 0.463

The instantaneous rate of growth is 463,000 peopleper year.

46. G{t) =m Go

Go + (m - G0)e-fc"" '

where G0 = 200;m = 10,000; and k = 0.00001.

10,000(200)(a)G(/-) - 200+(10)000_200)e-(oooool)(10'00°)t

G(t) =10,000

1 + 49e-°u

(b) G{t)= 10,000(1+ 49e-°u)-1G'(t) = -10,000(1 +49e-°H)-2(-4.9e-01')

49,000e-olt~ (l+49e-01')2

10,000 ^G(6)-l +49e-o«~359

49 000e-OG

(c) After 3 years, t = 36.

(d) After 7 years, t = 84.

G(84)_l +49e-8-4~J8J149 OOOe-84

(e) It increases for a while and then gradually decreases to 0.


47-Gw-G„+(,:^..)c-^whereG"=4oo;m = 5200; and k = 0.0001.

(5200)(400)(a) G(t) - 40Q +(520Q _400)e(_0.oooi)(52oo)f

(400)(5200)~ 400 + 4800e-°-52'

_ 5200~ 1 + 12e-°-52'

(b) G(t) = 5200(1 + 12e-°-52')"'G'{t) = -5200(1 + 12e-°-rj2')-2(-6.24e-°-52<)

_ 32,448e-052/~ (1 + 12e-°-52t)2

5200G(l) = ttzz « 039v ' l + 12e-°-52

32,448e-°-52G(1) - (1 +12e-0.52)2 ~ ZJZ

(c) G«)S1+^«M2081

(d) G(io)=1+^..a«48n34 448e-52

(e) It increases for a while and then gradually decreases to 0.

48. P(x) = 0.04e-4x

(a) P(0.5) = 0.04e-4(°-5>= 0.04e-'2

« 0.005

(b) P(l)=0.04e-4(,)= 0.04e"4

» 0.0007

(c) P(2) = 0.04e~4(2)= 0.04e-8« 0.000013

P'(rc) = 0.04(-4)e-4:,:= -0.16e-4:,:

(d) P'(0.5) = -0.16e-4(°5)= -0.16e"2

« -0.022

(e) P'(l) = -0.16e-4<l)= -0.16e"4

« -0.0029


(f) P'(2) = -0.16e-4(2>= -0.16e"8

« -0.000054

49. V{t) = 1100[1023e-°02415t + 1]~4

(a) V(240) = 1100[1023e-002415(24°) + l]"4« 3.857 cm3

4 I'W(b) V= ^Trr3, so r(V) = ? ^~6 V 47r

r(3.857)=:^^2«o.973 cm

(c) V(t) = 1100[1023e-°024154 + l]"4 = 0.5

[1023e -0.02415* I 11—4 _+ I]"4 =1

2200

(1023e-°02415t + l)4 = 2200

1023e-°024l5< + 1 = 22001/''

1023e-°02415( = 22001/4 - 1

2200l'4 - 1,-0.02415t _

1023

/22001/4 - 1-0.02415* = In ' ^^ -V 1023 )

02415 U\ 1023 )

50. P(t) = 0.00239e00957t

(a) P(25) = 0.00239e00957(25>« 0.026%

P(50) = 0.00239e00957(5°)« 0.286%

P(75) = 0.00239e°-0957<75)« 3.130%

(b) P'(t) = 0.00239e00957< (0.0957)= 0.000228723e00957'

P'(25) = 0.000228723e°-0957(25>« 0.0025%/year

P'(50) = 0.000228723e00957(5°)« 0.0274%/year

P'(75) = 0.000228723e°-0957(75>« 0.300%/year

:.,-{51. f/PP=l-U0.96)014(-1 + St;[l-(0

126*+900

(a) When t = 180, URR.« 0.589. The patient hasnot received adequate dialysis.

(b) When t = 240, URR « 0.690. The patienthas received adequate dialysis.

(c) DtURR= - {(ln0.96)(0.96)°-,4t-1(0.14)

•(-ln0.96)(0.96)014,-,(0.14)+126*+ 900

271

g6)0.1«-l]J

t =-0.02415 V 1023

The tumor has been growing for almost 18 years.

(d) As t goes to infinity, e~00241bt goes to zero,and V(t) = 1100[1023e-002415'+l]-4 goes to 1100cm3, which corresponds to a sphere with a radiusof 3j (noo) ^ 64cm jt makeg senge tnafc a tu_mor growing in a person's body reaches a maximum volume of this size.

(e) By the chain rule,

dV— = 1100(-4)[1023e-°024l5t + l]"5

214 months(126t + 900)(8)-8*(126)

(126J+900)2When t = 240, DtURR « 0.001. The URR is increasing instantaneously by 0.001 units per minutewhen t = 240 minutes.

The rate of increase is low, and it will take a significant increase in time on dialysis to increase URRsignificantly.

p(x) =0.001131e01268T

(a) p(25) = 0.001131e01268(25) « 0.027

(b) When p(x) = 1,

0.001131e012G8:c = 1

1

(1023)(,,-0.02415t X-0.02415)= 108,703.98[1023e-°02416f + i]-sc-o.o24i 5t

At t = 240, ~ « 0.282.at

At 240 months old, the tumor is increasing in volumeat the instantaneous rate of0.282 cm3/month.

52.

a0.12G8x _

0.001131

10.1268a: = In

0.001131

1 . 1x = In

0.1268 0.001131

54

This represents the year 2024.

272

(c) p'{x) = 0.001131e01268a:(0.1268)= 0.0001434108e01268x

p'(32) = 0.0001434108e01268(32)« 0.008

The marginal increase in the proportion per yearin 2002 is approximately 0.008.


53. M(t) = 3102e"„ -0.022(200-50)

(a) M(200) = 3102e"eor about 3 kilograms.

-0.022(t-5G)

54. L(0 = 589[l-e-0168(t+2G82)]

(a) Over time, the length of the average cutlassfish approaches 589 mm assymptotically.

(b) (0.95)589 = 589[1 - e-o.iG8(H-2.G82)j0.95 = 1 - e-°ir>8('+2-682)

ln(0.05) = -0.168(t + 2.682)t = 15.1497...» 15 years old

(c) L'(t) =589[-e-°-,68('+2-682)](-0.168)_ gg 952e~0168(t+2-682)

L'(4)=98^952e-°-168<4+2(382)w 32.2 mm/year

When a cutlassfish is 4 years old, it is growing inlength at a rate of 32.2 millimeters per year.

(d) *

2974.15 grams,

(b) As t gets very large, -e-°022(t-5G) goes tozero, e-e-° 022(t-r,0) goes to i5 ancj M(t) approaches3102 grams or about 3.1 kilograms.

(c) 80% of 3102 is 2481.6.

-In

In I In

2481.6 = 3102e"c

2481.6

-0.022(<-50)

3102

3102

_ _-0.022(t-5G)= e

(lnJM*L) =-0.022(* - 56)V 2481.6J v

t = - 022 V 2481.670.022

124 days

55.

(d) DtM(t) =3102e--OO22('-5C,A(-e-0O22(t-56))= 3102e-e~ °O22<t"50)(-e-0022('-5G))(-0.022)= 68.244e-e-OO22('"5G,e-0022(t-56)

When *= 200, DtM(t) « 2.75 g/day.

, . 3200(e)

J -0(R:<I-5S)=|A/(0 =3l02e-f

i^OO

Growth is initially rapid, then tapers off.

(0Day Weight Rate

50 991 24.88

100 2122 17.73

150 2734 7.60

200 2974 2.75

250 3059 0.94

300 3088 0.32

V20

Wi(t) = 509.7(1 - O^le"000181')W2(t) = 498.4(1 - O.SSge"000219')1-25

(a) Both W\ and W2 are strictly increasing functions, so they approach their maximum values as* approaches oo.

lim Wl(t) = lim 509.7(1-0.941e-°00,8U)= 509.7(1 - 0) = 509.7

0.1 _ e-0.00181t0.941

1239 «*

0.9(498.4) = 498.4(1 - O^e-000219')1'250.9 = (l-0.889e-000219')1-25

1 - 0.908

0.889

1095 « *

_ e-0.002l9<

lim W2(t) = lim 498.4(1-0.889e-0•002ly,),t—oo t—oo

= 498.4(1 -0),25= 498.4

So, the maximum values of W\ and W2 are 509.7kg and 498.4 kg respectively.

(b) 0.9(509.7) = 509.7(1 - 0.941e-°0018U)0.9 = 1 - 0.941e-°00,8U

,-0.00219t\1.25


Respectively, it will take the average beef cowabout 1239 days or 1095 days to reach 90% ofits maximum.

(c) W[{t) = (509.7)(-0.941)(-0.00181)e-°0018U» 0.868126e-°00181t

W{(750) « 0.868126e-000181(75°)« 0.22 kg/day

Wi(t) = (498.4)(1.25)(1 - 0.889e-°00219t)0-25• (-0.889)(-0.00219)e-°00219'

« 1.21292e-000219t(l - 0.889e-°00219t)0-25W^(750) « 1.12192e-000219(75°)

• (1 - o.889e-°00219(75°))0-25« 0.22 kg/day

Both functions yield a rate of change of about 0.22kg per day.

(d) Looking at the graph, the growth patternsof the two functions are very similar.

^2500

(e) The graphs of the rates of change of thetwo functions are also very similar.

2500

56. R(c) = 3.19(1.006c)

^=3.19(lnl.006)(1.006c)^at at

Uc= 180 and ft = 15, then

^ =3.19(lnl.006)(1.006180)(15)« 0.840

57. (a) Go = 0.00369,m = 1, fc = 3.5

1G(t) =

1 -|- ( 1 1^ *>-3.5(l)t1 ^ V0.00369 L)e W

1

1 + 270e"3-5*

58.

273

(b) G'(t) = -(1 + 270e-3-5')-2 •270e-35t(-3.5)

_ 945e-35'

G(l) =

(1 + 270e"3-5t)2

1

1 + 270e-3-5(D

0.109

945p-3.5(l)

1 ' [l+ 270e-3-5(D]2

« 0.341

The proportion is 0.109 and the rate of growth is0.341 per century.

<C> G(2>=l +27oU<*>« 0.802

Q45e-3.5(2)

1 ; [l+270e"3-5(2)]2

« 0.555


<d> G<3>=1+270U(3>» 0.993

945-3.5(2)

y} [1 + 270e-3-5(2)]2« 0.0256


(e) The rate of growth increases for a while andthen gradually decreases to 0.

H(N) = 1000(1 - e~hN), k = 0.1H' = -1000e-°1N(-0.1)

= 100e-0,Ar

(a) #'(10) = lOOe"01^«36.8

(b) //'(100) = lOOe"0-1*100)« .00454

(c) tf'(lOOO) = lOOe"0^1000*«0

(d) 100e~01N is always positive since powers ofe are never negative. This means that repetitionalways makes a habit stronger.

274

59. P(t) = 37.79(1.012)'

(a) P(10) = 37.79(1.021)10 « 46.5

So, the U.S. Latino-American population in 2010was approximately 46,500,000.

(b) P'{t) = 37.79(ln 1.021X1.021)'« 0.7854(1.021)*

P'(10)=0.7854(1.021)10« 0.967

The Latino-American population was increasingat the rate of 0.967 million/year at the end of theyear 2010.

60. A(t) = 500e-°-3"A'\t) =500(-0.31)e-°'31£

= -155e-03U

(a) A'(A) = -155e-°-31(4>= -155c"1-24« —44.9 grams per year

(b) A'{6) = -155e-°-31(6>= -155c"1-86« -24.1 grams per year

(c) A'(10) = -155e-°-31(10>= -155c"3-1« —6.98 grams per year

-0.3U _(d) lim A'(t) = lim -155e 0

As the number of years increases, the rate of changeapproaches zero.

(e) A'(t) will never equal zero, although as t increases, it approaches zero. Powers of e are alwayspositive.

61. Q(t) = CV(1 - e-^RC)

0 _ e-t/RC I _<a)Je =f =CK •(-*)= cv

-tine

(•Hi"- Y-e-t/RC~ R

(b) When G = 10"5 farads, R = 107 ohms,and V = 10 volts, after 200 seconds

r = jo e-20o/(io7io-5) ~ L35 x 10-7 ampsc 10*


62. t(r) = 218 + 31(0.933)'*

(a) i(55)=218 + 31(0.933)55« 218.7 sec

(b) t'(n) = (31 ln0.933)(0.933)"i'(55) = (311n0.933)(0.933)55

» -0.047

The record is decreasing by 0.047 seconds per yearat the end of 2005.

(c) As n -* oo, (0.933)" -> 0 andt(n) -> 218. If the estimate is correct, then thisis the least amount of time that it will ever take

a human to run a mile. At the end of 2005, the

world record stood at. 3:43.13, or 223.13 seconds.

63. T{h) = 80e-00000fi5/l

dT 55/1 (-0.(

-0.0052e-0000065/ldh

dt

l±L =goe-o.ooootis/, (_0.000065^dt \ at)

dhIf h = 1000 and — = 800, then

dt

— = -O.0052e-0-O00065(1000)(800)dt

« -3.90

The temperature is decreasing at 3.90degrees/hr.

4.5 Derivatives of LogarithmicFunctions

1. y = In (8a:)

dy_dx dx

8a:)

d

= &(ln 8 + In x)

8)+£<in x)

X

1

X

y = In (-4a;)

£ =sM-4*)]-;|M+>»<-*>ld , „ d , , v _ , —1 1

= — In 4 + — ln(-x) = 0 + = -dx dx -x x

•ffl-[5(I)+9]1/2 - Lower Moreland Township School ... 4.3 The Chain Rule Thus, q2 = 7500. P'(q) =...

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